shi20396_ch16

# shi20396_ch16 - Chapter 16 16-1 (a) 1 = 0, 2 = 120, a = 90,...

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Chapter 16 16-1 (a) θ 1 = 0°, θ 2 = 120°, θ a = 90°, sin θ a = 1, a = 5in Eq. (16-2): M f = 0 . 28 p a (1 . 5)(6) 1 Z 120° sin θ (6 5 cos θ ) d θ = 17 . 96 p a lbf · in Eq. (16-3): M N = p a . 5)(6)(5) 1 Z 120° sin 2 θ d θ = 56 . 87 p a lbf · in c = 2(5 cos 30 ) = 8 . 66 in Eq. (16-4): F = 56 . 87 p a 17 . 96 p a 8 . 66 = 4 . 49 p a p a = F / 4 . 49 = 500 / 4 . 49 = 111 . 4 psi for cw rotation Eq. (16-7): 500 = 56 . 87 p a + 17 . 96 p a 8 . 66 p a = 57 . 9 psi for ccw rotation A maximum pressure of 111 . 4 psioccurs on the RH shoe for cw rotation. Ans. (b) RH shoe : Eq. (16-6): T R = 0 . 28(111 . 4)(1 . 5)(6) 2 (cos 0 cos 120 ) 1 = 2530 lbf · in Ans. LH shoe : Eq. (16-6): T L = 0 . 28(57 . 9)(1 . 5)(6) 2 (cos 0 cos 120 ) 1 = 1310 lbf · in Ans. T total = 2530 + 1310 = 3840 lbf · in Ans. (c) Force vectors not to scale x y F y R y R x R F x F Secondary shoe 30 8 y x R x F y F x F R y R Primary shoe 30 8

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408 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design RH shoe : F x = 500 sin 30° = 250 lbf, F y = 500 cos 30° = 433 lbf Eqs. (16-8): A = µ 1 2 sin 2 θ 120 0 = 0 . 375, B = µ θ 2 1 4 sin 2 θ 2 π/ 3 rad 0 = 1 . 264 Eqs. (16-9): R x = 111 . 4(1 . 5)(6) 1 [0 . 375 0 . 28(1 . 264)] 250 =− 229 lbf R y = 111 . 4(1 . 5)(6) 1 [1 . 264 + 0 . 28(0 . 375)] 433 = 940 lbf R = [( 229) 2 + (940) 2 ] 1 / 2 = 967 lbf Ans. LH shoe : F x = 250 lbf, F y = 433 lbf Eqs. (16-10): R x = 57 . 9(1 . 5)(6) 1 [0 . 375 + 0 . 28(1 . 264)] 250 = 130 lbf R y = 57 . 9(1 . 5)(6) 1 [1 . 264 0 . 28(0 . 375)] 433 = 171 lbf R = [(130) 2 + (171) 2 ] 1 / 2 = 215 lbf Ans. 16-2 θ 1 = 15°, θ 2 = 105°, θ a = 90°, sin θ a = 1, a = 5in Eq. (16-2): M f = 0 . 28 p a (1 . 5)(6) 1 Z 105° 15° sin θ (6 5 cos θ ) d θ = 13 . 06 p a Eq. (16-3): M N = p a . 5)(6)(5) 1 Z 105° 15° sin 2 θ d θ = 46 . 59 p a c = 2(5 cos 30°) = 8 . 66 in Eq. (16-4): F = 46 . 59 p a 13 . 06 p a 8 . 66 = 3 . 872 p a RH shoe : p a = 500 / 3 . 872 = 129 . 1 psi on RH shoe for cw rotation Ans. Eq. (16-6): T R = 0 . 28(129 . 1)(1 . 5)(6 2 )(cos 15° cos 105°) 1 = 2391 lbf · in LH shoe : 500 = 46 . 59 p a + 13 . 06 p a 8 . 66 p a = 72 . 59 psi on LH shoe for ccw rotation Ans. T L = 0 . 28(72 . 59)(1 . 5)(6 2 )(cos 15° cos 105°) 1 = 1344 lbf · in T total = 2391 + 1344 = 3735 lbf · in Ans. Comparing this result with that of Prob. 16-1, a 2.7% reduction in torque is achieved by using 25% less braking material.
Chapter 16 409 16-3 Given: θ 1 = 0°, θ 2 = 120°, θ a = 90°, sin θ a = 1, a = R = 90 mm, f = 0 . 30, F = 1000 N = 1kN , r = 280 / 2 = 140 mm, counter-clockwise rotation. LH shoe : M f = fp a br sin θ a · r (1 cos θ 2 ) a 2 sin 2 θ 2 ¸ = 0 . 30 p a (0 . 030)(0 . 140) 1 · 0 . 140(1 cos 120 ) 0 . 090 2 sin 2 120° ¸ = 0 . 000 222 p a N · m M N = p a bra sin θ a · θ 2 2 1 4 sin 2 θ 2 ¸ = p a . 030)(0 . 140)(0 . 090) 1 · 120° 2 µ π 180 1 4 sin 2(120°) ¸ = 4 . 777(10 4 ) p a N · m c = 2 r cos µ 180 θ 2 2 = 2(0 . 090) cos 30 = 0 . 155 88 m F = 1 = p a · 4 . 777(10 4 ) 2 . 22(10 4 ) 0 . 155 88 ¸ = 1 . 64(10 3 ) p a p a = 1 / 1 . 64(10 3 ) = 610 kPa T L = a br 2 (cos θ 1 cos θ 2 ) sin θ a = 0 . 30(610)(10 3 )(0 . 030)(0 . 140 2 ) 1 [1 ( 0 . 5)] = 161 . 4N · m Ans. RH shoe : M f = 2 . 22(10 4 ) p a N · m M N = 4 . 77(10 4 ) p a N · m c = 0 . 155 88 m F = 1 = p a · 4 . 77(10 4 ) + 2 . 22(10 4 ) 0 . 155 88 ¸ = 4 . 49(10 3 ) p a p a = 1 4 . 49(10 3 ) = 222 . 8kPa Ans. T R = (222 . 8 / 610)(161 . 4) = 59 . 0N · m Ans .

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410 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 16-4 (a) Given: θ 1 = 10°, θ 2 = 75°, θ a = 75°, p a = 10 6 Pa, f = 0 . 24, b = 0 . 075 m (shoe width), a = 0 . 150 m, r = 0 . 200 m, d = 0 . 050 m, c = 0 . 165 m .
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## This note was uploaded on 04/30/2008 for the course MAE 4900 taught by Professor Khanna during the Fall '06 term at Missouri (Mizzou).

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shi20396_ch16 - Chapter 16 16-1 (a) 1 = 0, 2 = 120, a = 90,...

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